I would like to prove the existence and uniqueness of such a plane $P$, the truth is I can't think of a relatively simple way to prove it.
Let $\Lambda$ be a circle lying in $S$. Then there is a unique plane $P$ in $\mathbb{R}^3$ such that $P\cap S=\Lambda$. Recall from analytic geometry that $$P=\left\{(x_1,x_2,x_3):\beta_1 x_1+\beta_2 x_2+\beta_3 x_3-l=0\right\}$$ where $(\beta_1,\beta_2,\beta_3)$ is a vector orthogonal to $P$ and $l$ is some real number. Use this information to show that if $\Lambda$ contains the point $N$ then its projection on $\mathbb{C}$ is a straight line. Otherwise, $\Lambda$ projects onto a circle in $\mathbb{C}$.
My solution:
By the hint, exists a unique plane $P=\left\{(x_1,x_2,x_3)\in\mathbb{R}^3:\beta_1x_1+\beta_2x_2+\beta_3x_2-l=0\right\}$ such that $P\cap S=\Lambda$.\ Let $(x_1,x_2,x_3)\in \Lambda$ then $\varphi((x_1,x_2,x_3))=x+iy\in \mathbb{C}$ where $\displaystyle x=\frac{2x}{1-x_3},\ y=\frac{2y}{1-x_3}$ with $\varphi$ stereographic projection.
Moreover, for $z=x+iy$ it representation $\left(x_1,x_2,x_3\right)$ in $S$ is $\displaystyle(x_1,x_2,x_3)=\left(\frac{2x}{|z|^2+1},\frac{2y}{|z|^2+1},\frac{|z|^2-1}{|z|^2+1}\right)$. If this point belongs in $P$,
$$\beta_1\left(\frac{2x}{x_1^2+x_2^2+1}\right)+\beta_2\left(\frac{2y}{x_1^2+x_2^2+1}\right)+\beta_3\left(\frac{x_1^2+x_2^2-1}{x_1^2+x_2^2+1}\right)-l=0$$ equivalent to \begin{align}\label{ecuacion}(\beta_3-l)x^2+2\beta_1x+(\beta_3-l)y^2+2\beta_2y-\beta_3-l=0 \end{align} If $N=(0,0,1)\in \Lambda$ then $\beta_1\cdot 0+\beta_2 0+1\cdot \beta_3-l=0$, e.g. $\beta_3-l=0$, so, the equation has the form $$Ax+By+C=0$$ where $A=2\beta_1,\ B=2\beta_2,\ C=\beta_3-l$, general line equation.
If $N=(0,0,1)\not\in \Lambda$ then $\beta_3-l\neq 0$, so, the equation has the form $$Ax^2+By^2+Cx+Dy+E=0$$ where $A=B=\beta_3-l\neq 0,\ C=2\beta_1,\ D=2\beta_2,\ E=-\beta_3-l$, general circumference equation